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11111nata11111 [884]
3 years ago
10

Why is hydroelectric power considered to be an example of multiple transfers of energy?

Physics
1 answer:
Radda [10]3 years ago
7 0
Hydroelectric power is considered to be an example of multiple transfers of energy because potential energy when the water is in the reservoir turns into kinetic energy making the water move to the dam and then the dam moves into the turbine turning it to electric energy.  
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A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
3 years ago
Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod
aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force=\frac{Gm_{1}m_{2} }{r^{2}}

Coulombs's Law= \frac{Kq_{1}q_{2} }{r^{2}}

6 0
3 years ago
Read 2 more answers
A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the
shutvik [7]
Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.
6 0
3 years ago
Read 2 more answers
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature
shepuryov [24]

You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks.  Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.

For 'ideal' gases,

       (pressure) times (volume) is proportional to (temperature).

And if volume is constant, then

                 (pressure) is proportional to (temperature) .

So if you increase the temperature from 110K to 235K,
the pressure increases to  (235/110)  of where it started.

     (400 kPa) x (235/110)  =  854.55 kPa. (rounded)

Obviously, choice-b is the right one, but
I don't know where the .46 came from.
 
4 0
3 years ago
Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut
Harman [31]

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

4 0
3 years ago
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